Codeforces gym 101243 C - Desktop
2017-10-04 23:52
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链接
http://codeforces.com/gym/101243/题意
给一个宽高为w,h的矩阵,向里面塞2*2的小矩形,小矩形可以互相覆盖,但必须保证一个小矩形至少有一半是露在外面的,输出最多能摆的矩形数量和任一方案。分析
前两排可以以上图的方式摆放,共计(w - 1)个。
后几排以如上方式摆放,注意如果w是奇数是无法摆满的。
如果w为奇数,每两行需要添加如图所示的红矩形
交换w,h取较优答案
注意输出顺序必须按你的方案的摆放顺序,即后输出的在上面
#include <bits/stdc++.h> using namespace std; int main() { freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); vector<pii> ans; int h, w; scanf("%d %d", &h, &w); if (h == 1 || w == 1) { puts("0"); } else { bool flag = false; int answ = (w - 1 + (w / 2) * (h - 2)); if (w % 2 != 0) answ += (h - 2) / 2; int ansh = (h - 1 + (h / 2) * (w - 2)); if (h % 2 != 0) ansh += (w - 2) / 2; if (answ >= ansh) flag = false; else flag = true; if (flag) swap(h, w); for (int i = 1; i < w; ++i) { ans.push_back(pii(1, i)); } for (int i = 2; i < h; ++i) { for (int j = 1; j < w; j += 2) { ans.push_back(pii(i, j)); } } if (w % 2 != 0) { for (int i = 3; i < h; i += 2) { ans.push_back(pii(i, w - 1)); } } printf("%d\n", ans.size()); for (int i = ans.size() - 1; i >= 0; --i) { if (flag == false) printf("%d %d\n", ans[i].first, ans[i].second); else printf("%d %d\n", ans[i].second, h - ans[i].first + 1); } } return 0; }
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